0200 Sets and Classes

8/3/2019 0200 Sets and Classes

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Basic Set Theory

John L. Bell

I. Sets and Classes.

We distinguish between objectsand classes. Any collection of objects is

deemed to form a classwhich is uniquely determined by its elements. We

write

aA

to indicate that the object ais an element or memberof the class A. We

assume that every member of a class is an object. Lower-case letters a, b,c, x, y, z, will always denote objects, and later, sets.

Equalitybetween classes is governed by the Axiom of Extensionality:

Axiom A = B x[xAxB].

One class is said to be a subclassof another if every element of the

first is an element of the second. This relation between classes is denoted

by the symbol . Thus we make the

Definition AB x[xAxB].

Asubclass Bof a class Asuch that BAis called apropersubclass ofA.

Every property of objects determines a class. Suppose (x) is thegiven property of objects x; the class determined by this property is

denoted by

{x: (x)},

which we read as the class of all x such that(x). Churchs schemeis anaxiom guaranteeing that the class named in this way behaves in the

manner expected:

Axiom y[y {x: (x)} (y)].

Among classes we single out the universalclass V comprising all

objects and the emptyclass which has nomembers. Thus we make the

Definition V ={x: x = x} = {x: xx}.


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We shall sometimes write 0 for .A setis a class which is also an object. The purpose of a theory of

sets is to formulate existence principles which ensure the presence of

sufficiently many sets to enable mathematics to be done.

Russells Paradox shows that not every class can be a set. For

consider the class{x: x x} = R

(here x xstands for not x x). Suppose this class were a set r.Then itfollows from Churchs scheme that

rr rr,

a contradiction. Therefore Ris not a set.

In the present formulation of the theory of sets we quantify only

over objects, and notover classes in general. We do, on the other hand,

namemany classes and stateprincipleswhich apply to all classes.

Definitions

{a} =df {x: x = a}

{a1, , an} =df {x: x= a1 x =an}xA(x) df x[xA(x)]xA(x) df x[xA(x)]{xA:(x)} =df {x: xA(x)}

x, y, , zA df xA yA zAa1a2 an df a1a2 an-1 an

AB =df {x: xAxB}AB =df {x: xAxB}

A =df {x: xA}AB =df {x: xAxB}

Notice that , and satisfy the following laws ofBoolean algebra

A B = B A, A B = B A;A A = A, A A = A;

(AB) B = B, (AB) B = B

A (B C) =(A B) (A C), A (B C) =(A B) (A C),A A = V, A A =

At this point we make the simplifying assumption that everything

is a class.This is of course equivalent to asserting that every object is a

class, i.e., a set. It follows that the universal class Vis also the class of all

sets, that is,

XV Xis a set.


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Moreover, anyclass Ais now a classorfamilyof sets, and the sets in

Acan be unioned or intersected together to form a new class. Formally,

we make the

Definitions A =df {y: xA(yx)} A =df {y: xA(yx)}A and A are called the unionand intersectionofA, respectively. One

now easily proves the

Theorem = = V.

For any class B, the sets x Bthe subsetsofBform anotherclass called thepower classofB:

Definition PB =df {x: xB}.

We now introduce Zermelos axiomsfor set theory.

The axiom (or scheme) of separation tells us that whenever we

separate the elements of a setinto a subclass, the result is again a set.

That is, every subclass of a set is a set.Thus we have the

Axiom AaV A V,

or, equivalently, for any property (x),

aV {xa: (x)} V.

One easily derives from this the

Theorem VV.

Since the axiom of separation does not give us any sets to start

with, we need to assume that there is at least one set. But then it follows

from the axiom of separation that , which is clearly a subclass of everyclass, will also have to be a set. In that case we might as well assume the

axiom of the empty set:

Axiom V.

We also assume the axioms ofpairing, union, and power set:


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Axioms xy{x, y} VAV AV

AV PAV.


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II. Relations and Functions.

As we have seen, every property of objects determines a unique class.

Properties ofpairs of objects are called relations; does every relation

determine a class? This will be the case provided that each pair of objects

can itself be regarded as an object. Zermelos axioms in fact guarantee

this.

Since not all relations are symmetric, what we require is not the

unorderedpair {a, b} but some notion oforderedpair . An adequate

definition of ordered pair can be introduced by iterating the operation of

unordered pair in an asymmetric manner. Thus we make the

Definition =df {{a}, {a, b}}.

One then proves without much trouble the

Theorem = a = x b = y.

To dispel the impression that this definition has fallen out of a

clear blue sky, notice that as defined above is the set of initial

segmentsof the ordered set {a, b} in which aprecedes b.

The Cartesian productof two sets Aand Bis introduced by means

of the

Definition AB =df {: xAyB}.

Here the r.h.s. is an abbreviation for {z: xAyB[z =}: we shalluse similar abbreviations in the sequel. Notice that ABPP(AB),whence

A,BV ABV.

A (binary) relation may now be regarded as a class of ordered pairs.

For if (x, y) is a property of pairs x, yof objects, membership in the

class {: (x,y)} completely determines the pairs having the property. Since VVis evidently the largest class of ordered pairs, we make the

Definitions Rel[R] df RVV (Ris a relation)Ris a relation on A df RAA

xRy df Rdom R=df {x: y. xRy } (thedomainofR)

ran R=df {y: x. xRy} (the rangeofR)


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A single-valued relation is called a function. Thus we make the

further

Definitions !x(x) df yx[(x) x = y}

Fun[F] df Rel[F] xdom Fyz[xFyxFzy = z] (Fis afunction.)

IfF is a function, we define the function value F(x) or Fxso that

F(x) = Vifx dom F.This is convenient because VV, so to say that Fisdefined at xis just to say that F(x) V. Formally we make the

Definition F(x) =df y: Fun[F] xFy}.

One now proves

Cantors Theorem. There is no function Ffor which there is a set awithdom F= aand ran F = Pa.

Proof. Suppose Fun[F] , dom F = a V and ran F Pa. Defineb= {xa: x F(x)}. Then bPa. Ifb ran F, then b= F(c) for some ca.But then

cb cF(c) = b,

a contradiction. Therefore c ran F, so that ran FPa.

It will also be convenient to make the

Definitions R1 =df {: xRy } (theinverseofR)

R[A] =df {y: xA. xRy}R S =df {: y[xSyyRz]}1RA =df {: xAxRy}

F: AB dfFun[F] dom F = A ran F = B (Fis a functionfromA toB)Fis one-oneor injectiveor an injection

df xdom Fydom F[F(x) = F(y) x = y]F: ABis onto Bor surjectiveor a surjection df ran F = B

F: ABis bijective or a bijection df Fis one-one and onto B

BA =df {f: [f: AB]}.

Since BAP(AB), it follows that, ifA, BV, then BAV. Also,for any set a, and any class B, every elementfBais a functionf: aB.But what about the converse? Is every function F: Baan element ofBa? To ensure that this is the case we add the Axiom of Replacement:

1 Note here that, ifFand G are functions, then (F G)(x) =F(G(x)).


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Axiom Fun[F] dom FV ran FV.

From this one easily derives the required

Theorem Fun[F] dom FV FV.

The notion of ordered pair can be iterated to yield that ofordered n-

tuple: thus we make the

Definition (n 3) =df ,xn>

An =df {: x1, , xnA}.A subset ofAnis called an n-ary relation on A.

Theorem. = x1 = y1xn= yn.

Indexed sets. An indexingof a set Ais a functionf: IAfrom a set Icalled the index setonto A. One then usually writes:

aiforf(i), {ai: iI} for A, ii I

a for A, i

i I

a for A.


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III. Well-orderings and Ordinals.

Definitions. A (partial) ordering on a class A is a relation on Asatisfying, for all x, y, zA,

(i) xx (reflexivity)(ii) xy yz xz(transitivity)(iii) xy yx x= y(antisymmetry).

If in addition satisfies, for all x, yA,

(iv) xy xy(dichotomy),then is called a totalor linearordering ofA.

A strict linear ordering of A is a relation < on Asatisfying, for all x,

y, zA,

(a) xx (irreflexivity)

(b) x< y y< z x< z(transitivity)(c) x< y y< x x= y(trichotomy).

Linear orderings and strict linear orderings are interchangeable in view of

the fact that

is a linear ordering < (=df ) is a strict linear ordering,< is a strict linear ordering (=df< =) is a linear ordering.

If is a linear ordering, etc., of a set A,we shall call the pair A = ,A a

linearly ordered set, etc. The set A is called the underlying setofA.Weoccasionally identify Awith its underlying set A.

Definition. A well-ordering of a class Ais a linearordering ofAsuchthat

(a) each nonempty subset X A has a (necessarily unique) leastelement w.r.t. , i.e., an element aXsuch that ax for allxX2;

(b) for any aA, {xA: xa} is a set.A strict well-orderingofAis a strict linear ordering < ofAsuch that

(i) each nonemptysubset XAhas a (necessarily unique) minimalelement w.r.t.


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If < is a strict well-ordering ofAand XA, then a


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If C is the class of all partially ordered setsor even just the

subclass of linearly ordered setsit is not possible to define an order-

type operation on C explicitly. However, we show that when C is the class

W of well-ordered sets, the corresponding order-type operation can

actually be defined through the concept ofordinal number.

Thus we seek a class OWand a surjection F:W Osatisfying

XWYW[FX = FY X Y]XW[FX X].

The underlying sets of the members ofOwill be called ordinals. We now

define ordinals explicitly.

Definitions Xis transitive: Trans(X) df xX[xX]Xis an ordinal: Ord(X) df Xis a transitive set and the relation

|X=df{XX: xy} is a strict well-ordering ofX.

We use letters , , , , , to denote ordinals. Note that 0 = is anordinal.

Theorem. Each element of an ordinal is an ordinal.

Proof. Let be an ordinal and x. Then x, so, since | is a well-ordering of, |xis a well-ordering ofx. Ifzyx, then y becausex. Since | is a transitive relation on , it follows that zx. So xistransitive and hence an ordinal.

Definitions . For ordinals , we write

< for ; for ( < = ).

Let ,P be a partially ordered set. An initial segmentofPis a subset XofPsuch that

xXyP[yxyX];

it isproperifXP.Thus by an initial segment of an ordinal we mean asubset Xof such that X[ X];

Theorem. The only initial segments of an ordinal are the ordinal itself

and its initial segments.

Proof. Let Xbe an initial segment of an ordinal . IfX, then Xhas a .


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otherwise X because X is an initial segment of . HenceX ={: < } = = since .

Corollary. For ordinals , ,

.

Hence 0 for any .

Theorem. For each ordinal , .Proof. If , then since | is a strict ordering on . Inparticular, if then , and the contention follows.

Theorem. For ordinals , , exactly one of = , < , < holds.Proof. Put = . Then is an initial segment of, for ifyx

then x and x, whence y. Similarly, is an initial segment of.Hence

[ = or ] and [ = or ].

Accordingly there are four (actually three) possibilities:

= and = , in which case = ; = and , in which case ; = and , in which case ;

and , in which case = , contradicting .

One cant have = and (or ) since (and ); nor canone have and , for then , contradicting . Theresult follows.

Definition ORD ={x: Ord(x)}.

Theorem.The relation < (i. e., ) is a (strict) well-ordering ofORD.Proof. Most of this has been established above. It only remains to show

that any nonempty subset Xof ORDhas a


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hence an ordinal, and so aa, contradicting the fact that no ordinal canbe a member of itself.

Theorem. (i) For any ordinal , the least ordinal > is + 1 =df {}.(ii) For each set Xof ordinals, X is an ordinal which is the

least upper bound ofXw.r.t. , then and , whence {} , i.e., + 1 .

(ii) Put = X.To begin with, is an ordinal. For ifZ , then

Z for some X, and Z has a least element which is clearlyalso the least element ofZ. Thus | is a well-ordering of. Also, istransitive: for ifyx, then x for some X,whence y, andso y. Finally, we have for any X; if for all X, then for all X;so , whence . Thus is the least upper boundfor Xas claimed.

Theorem. Let and be ordinals and letfbe an order-isomorphism of, . So if, then

< f() = < , f() f() > .

Hence ran f, contradicting the fact that and, again, theassumption thatfis an isomorphism.

From the fact that < is a strict well-ordering of ORD we

immediately obtain the

Least ordinal principle. For any property (x),

() ![() [() ]],

as well as the

Principle of transfinite induction. For any property (x),


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[( < ()) ()] ().

Here the implication ( < ()) () is called the induction step.

Theorem. Principle of transfinite recursion. Suppose we are given

F: VV.Then

(i) (!g: V)


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completing the induction step and the proof of(i).

(ii) By uniqueness we have g= g for < , so that

G() = F(g) = F(g) = g().

Hence G = g, so that

G() = F(g) = F(G).

Definition An ordinal is a limitordinal, written Lim(), if it is neither 0nor of the form + 1.

Another form of transfinite induction. For any property (x),

[(0) [() ( + 1)] [[Lim() < ()] ()]] ().

This form of transfinite induction is easily derived from the original one.

Another form of transfinite recursion. Suppose given a0 V,F1, F2: VV. Then there is a unique G: ORDVsuch that

G(0) = a0, G( + 1) = F1(G()), G() = F(G) for limit .

To obtain G, apply the original principle of transfinite recursion to thefunction F: VVdefined by

if Fun[x] dom xORD

a0 ifx =F(x) =

F1(x()) if Fun[x] dom x= + 1

F2(x) if Fun[x] [ Lim() dom x= ]

We use this form of transfinite recursion to define the operations of

additionand multiplicationon ordinals as follows:

+ 0 = , + ( + 1) = ( + ) + 1, + = { + : < } for limit ;

.0 = 0, .( + 1) = . + , . = {.: < } for limit .


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Now we can prove the

Theorem. Each well-ordered set ,A


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all < . Hence , so that = . Therefore = , whence.

Next, we introduce the natural numbers.

Definition xis a natural number: N(x) df Ord(x) xLim().

Natural numbers are also called finite ordinals. An ordinal which is not

finite is called infinite.

The following is now easily proved:

Theorem (i) For any class A,

0 Ax[xAx+ 1 A]x[N(x) xA].(ii) If, are finite ordinals, so are + , ..

Notice that the natural numbers in order are 0, 1 = {0}, 2 = {0, 1},

3 = {0, 1, 2}, , n+ 1 = {0, 1, 2, , n}, .

Our penultimate axiom is the Axiom of Infinity:

Axiom There is an infinite ordinal.

Let us denote the leastinfinite ordinal by . Then we have the

Theorem. (i) is the least limit ordinal(ii) is the set of all finite ordinals.

Proof. (i) Evidently 0; if < , then is finite, so + 1 is also finite;hence + 1 . Thus is a limit ordinal. If is any limit ordinal, clearly must be infinite, so . This proves (i).(ii) If is finite, we cannot have because is a limit ordinal by (i).Therefore < , i.e., . Conversely, if, then < , so is finiteby definition of.

The axiom of infinity may be stated in the following three

equivalent forms:

The class of finite ordinals is a set; There is a limit ordinal; x[0 xy[yxy {y} x]].

Zermelo-Fraenkel set theory ZF is the axiomatic theory based on the

axioms of separation, empty set, pairing, union, power set, replacement,

and infinity.


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IV. Cardinal Numbers and the Axiom of Choice

We begin with the

Definitions XY there is a bijection between Xand Y.XY there is an injection ofXinto Y

XY XYXY.

It is easily seen that is an equivalence relation on V. XY is read Xand Yare equipollent.

For the present we shall assume that we have a function

|.|: VVcalled the cardinality functionsatisfying the following axiom ofcardinalities:

Axiom XVYV[|X| = |Y| XY]

|X| will be called the cardinalityof X. Later on we will show (once we

have the axiom of choice) that the function |.| can actually be defined,

and the axiom of cardinalities derived from the other axioms.

Definition. CARD =df{|X|: XV}.

Members ofCARDare called cardinals. We shall use letters m, n, p, etc.

for cardinals.

Definitions X+ Y =df(X {0}) (Y {1})

i

i I

A

=df { }ii I

A i

Ifm = |X|, n = |Y|, we define

m + n =df|X+ Y|, m.n =df|X Y|, |, mn =df|XY|,

m c n =dfXY, m


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For a natural number k,

k.m =dfm + m + + m, mk= m.m m (both ktimes).

Theorem. m+ n = n + m, m.n = n.m(m + n) + p = m + (n +p), (m.n).p = m.(n.p)

(m + n).p = m.p + n.p

m c m n c n m + n c m + nif mi m for all iI, then

i I mi c |I|.m

Theorem (Cantor). |A| |PA|.

Axiom of Choice. XV(f: XX)xX[xf(x) x].

Such a functionfis called a choice functionon X.

Definition. A nonempty partially ordered set P is said to be inductive if

each nonempty linearly ordered subset (also called a chain) has an upper

bound in P. An element ofP is maximal if it is not strictly smaller than

any element ofP.

Using the axiom of choice we now prove

Zorns Lemma. Each inductive set has a maximal element.

Proof. Let ,A be a partially ordered set, let hbe a choice function forPAand let

B= {XA: Xhas a strict upper bound}.

(Here by a strict upper bound for a subset X of P we mean an upper

bound not in X.)Definef: BAby

f(X) = h({x: xis a strict upper bound for X}).

Now let ebe any set which is not a member ofA. By transfinite recursion

define F: ORDVby

f({F(): < }) if {F(): < } BF() =

e if {F(): < } B


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Observe that ran FA {e}. If ran FA, then {F(): < } B for all .Hence, if < , then F() F(). In that case F1is a function from AontoORD, so that ORD would be a set by the axiom of replacement, a

contradiction.

It follows that F takes the value esomewhere. Let 0 be the first

ordinal at which it does. If < 0, then F() A, so {F(): < } B andthus F() is a strict upper bound for {F(): < }. Therefore

< < 0F() < F().

Hence F0 is an order-preserving injection of0 into ,A . In that case{F(): < 0} is a linearly ordered subset ofAand so has an upper bounda. Ifa< b, then bwould be a strictupper bound for {F(): < 0}, so that{F(): < 0} B and F(0) e, a contradiction. Therefore ais a maximalelement ofP.

Corollary. The Well-Ordering Theorem. Every set can be well-ordered,

and is therefore equipollent to an ordinal.

Proof. Let B be the set of all pairs ,B with BAand a well-orderingofB.Then B; partially order B by

,B ', 'B BB, = Band Bis an initial segment ofB.

We claim that B, is inductive. For let C = {{ ,i iB : iI} be any

chain in B. Then = ii I

is easily seen to be a linear ordering on B =

i

i I

B ;we show that it is a well-ordering. Let Xbe a nonempty subset of

B.Then XBi for some iIand so XBi has a least element, a,say, w.r.t i . IfxX, then, since C is a chain, there is j Isuch thatBi {x} Bj. Ifx< a, then xXBisince Biis an initial segment ofBj. This contradicts the choice ofa, so a x and therefore a is the leastelement ofX. Accordingly well-orders B.

Accordingly ,B is an upper bound for C and therefore B, is

inductive. Consequently, Zorns lemma applies that it has a maximalelement ,D . We claim that D = A, thereby proving the result. If, on thecontrary, bAD, then we define on D= D {b} by

D= D = ; x


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Clearly ', 'D B is strictly greater than ,D w.r.t. , contradicting

the maximality of the latter.

Corollary. Let m and n be cardinals. Then

(i) Trichotomy: m


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Let m = |A| and (Lemma 2) let BAsatisfy B. Then (Lemma 3)there is a bijectionf0: BBB. Let F be the set of pairs whereB X A and f is a bijection between X and X X such that f0 f.Partially order F by stipulating that

XXandff.

Then F, is clearly inductive and hence by Zorns Lemma has a

maximal element . We show that |C| = m, thereby proving the

theorem.

Suppose on the contrary that |C| c |C| = n.

Accordingly there is a subset Y A C such that |Y| = n; putZ = CY. We show that there is a bijection h: Z ZZsuch that gh.For we have

ZZ =(C C) (CY) (YC) (YY),

and the sets on the r.h.s. are disjoint. Since CY, we have

|CY| = |YC| = |YY| = n2 = n,

so that

|(C Y) (YC) (YY)| = 3.n = n.

Thus there is a bijection gofYonto (CY) (YC) (YY). So themap hofZinto ZZdefined by hC= gand hY= gis a bijection andgh.

But this contradicts the maximality of . therefore |C|


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Corollary. Ifm, n are cardinals 0, of at least one is infinite, thenm.n = m + n = larger ofm, n.

Proof. Suppose m cn.Then

n c m.n c n2 = n; n c m + n c 2.n c n2 = n. #

Now we can make the

Definition. For any set A,

|A| = leastsuch that A.

Then we have the

Theorem. (i) AV A |A|

(ii) Card(x) Ord(x) |x| = x Ord(x) (e.g. = + 1). Then || || = , contradicting thefact that , as a finite ordinal, must be a cardinal. So is an infinitecardinal. It must be the least such because every smaller cardinal is

finite. #

Theorem. For each set A of cardinals there is a cardinal m which

exceeds every member ofA.

Proof. Take m = |P(A)|. #


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By transfinite recursion we define the cardinal ! (aleph-) foreach ordinal as follows:

! = least cardinal m such that m and m {!: < }.

Theorem. (i) Card(!) !!0 = ( < !< !).(ii) (Card(m) < m) m = !.

Proof. Most of(i) follows immediately from the definition of!. Only the

last part requires proof. Thus suppose < . Then ! {!: < }, so afortiori ! {!: < }. But ! is the least infinite cardinal not in{!: < }, so !< !.

To prove (ii), we first observe that, since the map ! is

injective, a straightforward application of the axiom of replacement

shows that {!

: ORD} is not a set. So, given an infinite cardinal m,there must be an for which ! m. Thus m !. If m = ! we arethrough. If m < ! then since ! is the least infinite cardinal not in

{!: < }, we must have m {!: < }. So m = ! for some < . #

Accordingly the alephs form an enumeration of all the infinite cardinals.


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Problems

1. Is VV= V?2. (i) Find a bijection between ABCand (AB)C.

(ii) What are A, A{}, A?(iii) Find a necessary and sufficient condition on A, Bfor AB BA.(iv) Show from first principles that AAA for any set Awith at least

two elements.

(v) Show that {x: xx} V, and generalize.

3. For each set Awrite 1A for the identity map on A.

(i) Letf: ABwith A. Show thatfis injective iff there is amap r: BAsuch that rf= 1A.. When is r unique?

(ii) Let f: ABand suppose that there is a map s: B Asuchthatf s =1B: sis called a sectionoff. Show that, in that case,f

is onto. Does everyonto map have a section?

4. Letf: AB. Define : PBPAby (X) =f1[X] for XB. Show thatfis injective(surjective) iff is surjective (injective).

5. Define theprojectionmaps 1: ABA, 2: ABBby 1() =a, 2() = bfor aA, bB.

(i) Given maps f: C A, g: C B, find the unique maph: CABfor which the diagram

C

f h g

A 1AB 2 B

commutes (i.e. 1h=f, 2h= g).(ii) Suppose that 1: D A, 2: D B, are such that, for any

f: C A, g: C B, there is a unique h: C D such that thediagram

C

f h g

A 1 D 2 B

commutes. Show that there is a uniquebijection i: DABsuchthat the diagram


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D

1 i 2

A 1AB 2 B

commutes. What does this tell us about AB?6. Let {Xi: i I} be a family of subsets of a set Band letf: A B. Showthatf1[ i

i I

X ]=

i I f1[Xi] andf1[ i

i I

X ]=

1[ ]ii I

f X

. If {Yi: iI} is a family

of subsets ofA, show that f[ ii I

X ]= [ ]i

i I

f X . Does f[ i

i I

X ]=

i I f[Xi]

always?

7. Let {Aij: IJ } be a family of sets. Using the axiom of choice,prove the distributive law

( )I

ij if i

i I j J i I f J

A A

=

Deduce the dual formula

( )I

ij if i

i I j J i I f J

A A

= .

Show, conversely, that the distributive law implies the axiom of choice.

8. Letf: XEbe an injection of a set Xinto a subset EX.Define D ={yX: nxXE. y = fnx}, wherefnx is defined recursively byf0x = x,

fn+1x = f(fnx). Let h: XEbe the map defined by hD = fD, hXD=1XD.Show that h is bijective, and deduce the Schrder-Bernstein theorem: if

ABand BA, then AB.9. For a natural number n> 0, an n-familyis a family Aof sets with |A|

= nfor each AA.. Withoutusing the axiom of choice, prove that (i) if forsome k> 0 every kn-family has a choice function, then so does every n-

family; (ii) if every 2-family has a choice function, then so does every 4-

family.

10. A set Ais transfiniteifAand Dedekind infiniteif there is a proper

subset BofA for which BA. Withoutusing the axiom of choice, showthat, if A is infinite, then PPA is transfinite. Again without using the

axiom of choice, show that transfiniteness is equivalent to Dedekindinfiniteness, and that the former implies infiniteness. Assuming the

axiom of choice, show that infiniteness implies transfiniteness.

11. A set Ais said to be countableif |A| !0. Show that: (i) the union oftwo countable sets is countable; (ii) assuming the axiom of choice, the

union of a countable family of countable sets is countable; (iii) the set of

all finite subsets and of all finite sequences of members of a countable

set is countable; (iv) a set is countable iff it is the union of a chain of


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finite sets; (v) there is an uncountable set which is the union of a chain

of countable sets.

12.The transitive closure of a set A is the least transitive set which

includes A. Show that any set has a transitive closure.

13. A partially ordered set ,A is said to be completeif every subset of

Ahas a least upper bound and a greatest lower bound. Letfbe an orderpreserving map of a complete partially ordered set Ainto itself. Show that

fhas afixed point, i.e. there is aAfor whichf(a) = a.14. Let C be the set of all continuous maps of the set of real numbers

into itself. What is |C|? What is ||? Are they the same?

15. A family Fof subsets of a set Ais said to be localif, for any subset X

ofA, XF iff every finite subset ofXis in F. Using Zorns lemma, showthat every local family has a maximal member w.r.t. . Conversely, showthat this principle implies Zorns lemma.16. Derive the axiom of choice directly from Zorns lemma.17. Use Zorns lemma to give a direct proof of the comparability

principle: for any sets A, B, either ABor BA.

18. A family U of subsets of a set A is called an ultrafilter on A if

(a) U, (b) X, Y U X Y U, (c) for all X A, either X U orA XU.

(i) Let U be an ultrafilter on A and suppose that X U andXYA. Show that YU..

(ii) Show that for any a A the family {X A: a X } is anultrafilter on A. An ultrafilter of this form is called

principal.(iii) A family F of subsets ofA is said to be neighbourlyif for

finite subfamily of F has a nonempty intersection. A

neighbourly family F is said to be maximal if the only

neighbourly family which includes F is F itself. Show that

ultrafilters and maximal neighbourly families coincide.

(iv) Let Fbe a neighbourly family of subsets ofA. Use Zorns

lemma to show that F is included a maximal neighbourly

family, and hence an ultrafilter on A.

(v) Suppose that A is infinite. Show that the family ofcofinite(i.e., complement of finite) subsets ofAis neighbourly, and

deduce the existence of nonprincipal ultrafilters on A.

19.The axiom of foundationis the assertion xyx[xy}.Show that the axiom of foundation implies that

(*) there does not exist an infinite descending sequence of sets

x0x1 x2 xn


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Deduce that, if the axiom of foundation holds, there can be no nonempty

class A such that A A = A. Show also that, assuming the axiom ofchoice, (*) implies the axiom of foundation.

20. Define the sets R recursively by R = {x:

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0200 Sets and Classes